10. An aqueous solution is prepared by diluting 3.30 mL acetone, CH3COCH3, (d=0.789 g/mL) with water to a final volumeof 75.0 mL. The density of the solution is 0.993 g/mL. What is the molarity and molality of acetone in this solution?

To find the molarity and molality of acetone in the solution, we need to calculate the moles of acetone present.

First, let's calculate the mass of the acetone used in the solution. We'll assume that the density of acetone doesn't change during dilution.

Mass of acetone = volume of acetone × density of acetone
= 3.30 mL × 0.789 g/mL
= 2.6007 g

Next, let's calculate the mass of the solution.

Mass of solution = volume of solution × density of solution
= 75.0 mL × 0.993 g/mL
= 74.475 g

Now, we can calculate the moles of acetone.

Moles of acetone = mass of acetone / molar mass of acetone

The molar mass of acetone (CH3COCH3) can be calculated by adding the atomic masses of its constituent elements:

C: 12.01 g/mol × 3 = 36.03 g/mol
H: 1.01 g/mol × 6 = 6.06 g/mol
O: 16.00 g/mol × 1 = 16.00 g/mol

Molar mass of acetone = 36.03 g/mol + 6.06 g/mol + 16.00 g/mol = 58.09 g/mol

So, moles of acetone = 2.6007 g / 58.09 g/mol = 0.0447 mol

Now, let's calculate the molarity and molality of acetone in the solution.

Molarity (M) is defined as moles of solute divided by the volume of the solution in liters.

Molarity = moles of acetone / volume of solution in liters
= 0.0447 mol / (75.0 mL/1000)
= 0.596 M

Molality (m) is defined as moles of solute divided by the mass of the solvent in kilograms.

Molality = moles of acetone / mass of water in kg
= 0.0447 mol / (74.475 g/1000)
= 0.601 mol/kg

So, the molarity of acetone in the solution is 0.596 M, and the molality is 0.601 mol/kg.

To find the molarity of acetone in the solution, we first need to calculate the moles of acetone present.

Step 1: Calculate the mass of acetone.
Mass of acetone = volume of acetone × density of acetone
Mass of acetone = 3.30 mL × 0.789 g/mL
Mass of acetone = 2.6007 g

Step 2: Calculate the moles of acetone using the molar mass of acetone.
Molar mass of acetone (CH3COCH3) = (12.01 g/mol × 2) + (1.01 g/mol × 6) + (16.00 g/mol × 1) = 58.08 g/mol
Moles of acetone = mass of acetone / molar mass of acetone
Moles of acetone = 2.6007 g / 58.08 g/mol
Moles of acetone = 0.0448 mol

Step 3: Calculate the molarity of acetone.
Molarity (M) = moles of solute / volume of solution (in liters)
Volume of solution = 75.0 mL = 0.075 L
Molarity of acetone = 0.0448 mol / 0.075 L
Molarity of acetone = 0.597 M

To find the molality of acetone in the solution, we need to calculate the moles of acetone and the mass of water.

Step 4: Calculate the mass of water.
Mass of water = (mass of solution - mass of acetone)
Mass of solution = volume of solution × density of solution
Mass of solution = 75.0 mL × 0.993 g/mL
Mass of solution = 74.475 g

Mass of water = 74.475 g - 2.6007 g
Mass of water = 71.8743 g

Step 5: Calculate the molality of acetone.
Molality (m) = moles of solute / mass of solvent (in kg)
Mass of water (in kg) = mass of water / 1000
Mass of water (in kg) = 71.8743 g / 1000 g/kg
Mass of water (in kg) = 0.0718743 kg

Molality of acetone = 0.0448 mol / 0.0718743 kg
Molality of acetone = 0.622 m

Therefore, the molarity of acetone in the solution is 0.597 M, and the molality of acetone in the solution is 0.622 m.

Molarity= massAcetone/formulamass *1/.0033

massAcetone= density*volume

calculate molarity.

Finally, molarity= massAcetone/formulamass*1/(.075-.0033)